1 / 13

Continuous Stirred Tank Reactor

Continuous Stirred Tank Reactor. Problem statement. A chemical reaction takes place in a series of four continuous stirred tank reactors arranged as shown in Fig. 100 lit/hr. 100 lit/hr. 1000 lit/hr. C A0 =1 mol/lit. V 4 C A4 K 4. V 1 C A1 K 1. V 3 C A3 K 3. V 2 C A2 K 2.

acton
Download Presentation

Continuous Stirred Tank Reactor

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Continuous Stirred Tank Reactor

  2. Problem statement A chemical reaction takes place in a series of four continuous stirred tank reactors arranged as shown in Fig

  3. 100 lit/hr 100 lit/hr 1000 lit/hr CA0=1 mol/lit V4 CA4 K4 V1 CA1 K1 V3 CA3 K3 V2 CA2 K2 1000 lit/hr CA1 CA2 CA3 CA4

  4. The chemical reaction is a first order irreversible reaction of the type- A B • The value of the rate constant ki, is different in each reactor. Also, the volume of each reactor Vi is different k

  5. Assumptions: • The system is steady state and unsteady state. • The reactions are in liquid phase. • There is no change in volume or density of the liquid.

  6. Solution

  7. Material balance continued:

  8. Using MATLAB for steady state results function f=fourcstrsteady(x) f=zeros(4,1); %defining constants CA0=1; V1=1000; K1=0.1; %data from table V2=1500; K2=0.2; V3=100; K3=0.4; V4=500; K4=0.3; xa=x(1);xb=x(2);xc=x(3);xd=x(4); %material balance equations: f(1)=(1000*CA0)-(1000*xa)-(V1*K1*xa); f(2)=(1000*xa)+(100*xc)-(1100*xb)-(V2*K2*xb); f(3)=(1100*xb)+(100*xd)-(1200*xc)-(V3*K3*xc); f(4)=(1100*xc)-(1100*xd)-(V4*K4*xd);

  9. Running the following displays the steady state concentrations in the tanks: clc clear all x0=[0,0,0,0]; %initial values x=fsolve(@fourcstrsteady, x0) %fsolve to solve the steadystate

  10. MATLAB for unsteady state results function f=fourcstr(t,x) f=zeros(4,1); %defining constants CA0=1; V1=1000; K1=0.1;%data from the table given V2=1500; K2=0.2;%data from the table given V3=100; K3=0.4;%data from the table given V4=500; K4=0.3;%data from the table given xa=x(1);xb=x(2);xc=x(3);xd=x(4); %defining the differential equations %material balance equations assuming unsteady state f(1)=(1000*CA0)-(1000*xa)-(V1*K1*xa); f(2)=(1000*xa)+(100*xc)-(1100*xb)-(V2*K2*xb); f(3)=(1100*xb)+(100*xd)-(1200*xc)-(V3*K3*xc); f(4)=(1100*xc)-(1100*xd)-(V4*K4*xd);

  11. Running the following code in MATLAB yields the plot depicting the variation of Concentration in each tank: clc clear all x0=[1;0;0;0]; %defining the initial values. [t,x]=ode45(@fourcstr, [0 0.1], x0); %ode45 to solve the unsteady state figure; plot(t,x); %plot function %labelling x and y axes xlabel('time t(hrs)'); ylabel('concentration c(t)');

  12. Steady state result predicted : At steady state, the concentration in tanks 1,2,3 and 4 as predicted by the programme: [CA1 CA2 CA3 CA4]= [0.9091 0.6969 0.6654 0.5856]

  13. Unsteady state results The following variation is predicted with respected to time

More Related